Get Ready for Grade 8panchbhaya.weebly.com/uploads/1/3/7/0/13701351/pom9-solutions-chapter... · Chapter 5 Get Ready Chapter 5 Get Ready Question 1 Page 236 a) 3 4 − is not equivalent

Chapter 5 Modelling With Graphs Chapter 5 Get Ready Chapter 5 Get Ready Question 1 Page 236

a) 34− is not equivalent to the others, since it is negative. All of the rest are positive.

b) 52

is not equivalent to the others, since it is positive. All of the rest are negative.

c) 12−−

is not equivalent to the others, since it is positive. All of the rest are negative.

Chapter 5 Get Ready Question 2 Page 236

a) 2 0 45

.= b) 7 0 710

.− = −

c) 35 0 87540

.−= − d) 12 2 4

5.−

= −

Chapter 5 Get Ready Question 3 Page 236

a) 39 3

− = −1 b) 15 3

10 2− −

=

c) 12 148 4

−=

− d) 30 5

12 2=

− −

Chapter 5 Get Ready Question 4 Page 237 a) 5:20 = 1:4 b) 12:96 = 1:8 c) 12:14 = 6:7 d) 40:850 = 4:85 Chapter 5 Get Ready Question 5 Page 237 7 120 84

10× =

In a group of 120 people, 84 would prefer Fresh toothpaste.

MHR • Principles of Mathematics 9 Solutions 281

Chapter 5 Get Ready Question 6 Page 237 12 160 6430

× =

A person who is 160 cm tall is about 64 inches tall. Chapter 5 Get Ready Question 7 Page 237

Chapter 5 Get Ready Question 8 Page 237 Bag Nitrogen (kg) Phosphorus (kg) Potassium (kg)a) 10-kg 20:4:8 2 0.4 0.8 b) 25-kg 21:7:7 5.25 1.75 1.75 c) 50-kg 15:5:3 7.5 2.5 1.5 d) 20-kg 10:6:4 2 1.2 0.8

282 MHR • Principles of Mathematics 9 Solutions

Chapter 5 Section 1: Direct Variation Chapter 5 Section 1 Question 1 Page 242

a) 2803 5

80

k.

=

=

b) 355

7

k =

=

c) 5005

100

k =

=

Chapter 5 Section 1 Question 2 Page 243

a) 4500200

22 5

22 5

k

.

C .

=

=

= s

b) The constant of variation represents the cost for each meter of sidewalk. c) ( )70022 5

15 750C .=

=

The cost of a 700-m sidewalk is $15 750.


Chapter 5 Section 1 Question 3 Page 243 a) b) c) 8p t= Chapter 5 Section 1 Question 4 Page 243 a) b) c) 1 5c . a=


Chapter 5 Section 1 Question 5 Page 243 a) To calculate the cost of parking, multiply the time parked, in hours, by $2.75. The cost c, in dollars, of parking, varies directly with the time, t, in hours, for which the car is parked. b) 2 75c .= t

2 7519 25

c ..

=

=

c) Answers will vary. The cost is a little under $3 per hour, so 7 h should cost about $20. d) ( )7

It costs $19.25 for 7 h of parking. Chapter 5 Section 1 Question 6 Page 243 a) To calculate the cost C, of oranges, multiply the mass r, in kilograms, of oranges, by $2.25.

b) 4 502

2 25

2 25

.k

.

C .

=

=

= r

2 2567 50

C ..

=

=

The constant of variation represents the constant average cost, $2.25/kg. c) ( )30

It costs $67.50 to buy 30 kg of oranges.


Chapter 5 Section 1 Question 7 Page 244 a)

b) 50163 125

3 125

k

.

A .

=

=

= t

c) ( ) 3 125

75A .=

=

h

24

Tania would have raised $75 by staying awake for 24 h. Chapter 5 Section 1 Question 8 Page 244 a) 9 5P .= b) ( )1 5 9 5

14 25T . . h

. h=

=

c) 10P h=

( )1 5 1015

T . hh

=

=

Chapter 5 Section 1 Question 9 Page 244 a) This relationship is a direct variation because the price of the sugar varies directly with the amount of sugar that is bought. b) c) The graph shows that if the price increases to $1.49 for 0.5 kg (or $2.98/kg), the graph becomes steeper.


Chapter 5 Section 1 Question 10 Page 244 Answers will vary. Sample answers are shown. a) A cyclist travels 20 m in 2 s. b) A car is parked for 8 h. The cost of parking is $4. Chapter 5 Section 1 Question 11 Page 244 The time given is for the round trip from the bat to the object and back to the bat. In order to find the distance to the object, the distance must be divided by 2.

1 3422171

d t

t

= ×

=



a) 5004

125

125

k

V t

=

=

=

V is the volume of the water, in litres, and t is the time, in minutes. The constant of variation represents the rate of increase of the volume, 125 L/min. b) c) ( ) 20125

2500 LV =

= There are 2500 L of water in the pool after 20 min. d) 125

115 000 125125

115 000

592

120

tt

t

=

=

= It takes 920 min to fill a pool that holds 115 000 L of water.

e) 4004

100

100

k

V t

=

=

=

The graph would still increase to the right, but less steeply. It would take longer to fill the pool.


Chapter 5 Section 1 Question 13 Page 245 a) The freezing point depends on the salt content. The salt content is the independent variable. b) Let F represent the freezing point, in degrees Celsius, and s represent the salt content, as a percent.

3 521 75

1 75

.k

.

F . s

=−

= −

= −

c) ( )11 75

1 75F .

.= −

= − Water with a salt content of 1% will freeze at –1.75ºC. d) 1 75

3 1 751 75 1 75

3−

1 7

. s. s

. .. s

= −− −

=−−

To freeze at –3ºC, water must have a salt content of about 1.7%. Chapter 5 Section 1 Question 14 Page 245 Let k represent the number of kilometres, and m represent the number of miles.

0 620 62

0 62 0 621

0 62

m . km .. .

m k.

=

=

=

k

An equation to convert miles to kilometres is k = 1.61m. Chapter 5 Section 1 Question 15 Page 245 The circumference varies directly with the diameter. The constant of variation is π.


Chapter 5 Section 1 Question 16 Page 245 From 1 to 100, there are 19 disks that contain a 3: 3, 13, 23, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 43, 53, 63, 73, 83, and 93. So, the probability that a disk contains a 3 is 19 0 19100

19

.

%

=

=

Chapter 5 Section 1 Question 17 Page 245 The greatest possible number is 65 423. The least possible number is 23 465. The difference is 65 423 – 23 465 = 41 958.


Chapter 5 Section 2 Partial Variation Chapter 5 Section 2 Question 1 Page 250 a) This is direct variation. The equation is of the form y kx= . b) This is partial variation. The equation is of the form y mx b= + . c) This is partial variation. The equation is of the form y mx b= + . d) This is direct variation. The equation is of the form y kx= . Chapter 5 Section 2 Question 2 Page 250 a) b) The initial value of y is 5. The constant of variation is 5. c) 5 5y x= + d) e) The graph is a straight line that intersects the y-axis at (0, 5). The y-values increase by 5 as the x-values increase by 1.


Chapter 5 Section 2 Question 3 Page 251 a) b) The initial value of y is –2. The constant of variation is 5. c) 5 2y x= − d) e) The graph is a straight line that intersects the y-axis at (0, –2). The y-values increase by 5 as the x-values increase by 1.


Chapter 5 Section 2 Question 4 Page 251 a) The fixed cost is $7.00. The variable cost is $1.50 times the number of toppings. b) 1 50 7 00C . n .= + c) ( )1 50 7 00

7 50 7 0014 50

5C . .. ..

= +

= +=

The cost of a small pizza with five toppings is $14.50. Chapter 5 Section 2 Question 5 Page 251 a) The fixed cost is $250. The variable cost is $4 times the number of students. b) 4 250C n= + c) ( )4 250

100 250350

25C = +

= +=

The total cost for 25 students is $350. Chapter 5 Section 2 Question 6 Page 251 a) b) Membership A is a direct variation. Membership B is a partial variation. c) Membership A: C represents the cost, and n represents the number of visits. 4C n= Membership B: 2 12C n= + d) Membership A is cheaper when fewer than six visits are made. Membership B is cheaper when more than six visits are made. They cost the same when six visits are made.


Chapter 5 Section 2 Question 7 Page 252 a) The fixed cost is $100 and could represent, for example, the cost of paper, ink, and overhead. b) From the table, it costs $20 over the fixed cost to print 100 flyers, so the variable cost to print one flyer is $20 ÷ 100 or $0.20. c) 0 20 100C . n= + d) ( )0 20 100

200100100

00

0

3

C .= +

= +=

It costs $300 to produce 1000 flyers. e) 0 20 100

280 100 0 20 100 100180 0 20

180 0 200 20 0

280

20900

. n

. n

. n. n

. .n

= +− = + −

=

=

=

900 flyers can be produced for $280. Chapter 5 Section 2 Question 8 Page 252 a) , where T is the number of toothpicks and n is the diagram number. This is a partial variation because it is of the form

2 1T n= +y mx b= + .

b) ( )2

4120 1T = +

=


Chapter 5 Section 2 Question 9 Page 252 a) , where P is the pressure, in kilopascals, and d is the depth below the lake’s surface, in metres.

10 13 102 4P . d .= +

b) 10 13 102 4

400 102 4 10 13 102 4 102 4297 6 10 13297 6 10 1310 13 10 13

004

9 42

. d .. . d . .. . d. . d

. .. d

= +− = + −

=

=

The danger from narcossis begins at about 29 m of depth. Chapter 5 Section 2 Question 10 Page 252 Answers will vary. A sample answer is shown. A plumber comes to your house to repair a leak. It costs $30 for a service call, and $10 for each hour it takes to complete the job.



b) The average rate of descent is 7500 80002− , or –250 m/min.

c) , where H is the height above ground, in metres, and t is the time, in minutes.

250 8000H t= − +

Chapter 5 Section 2 Question 12 Page 253 Solutions for the Achievement Checks are shown in the Teacher's Resource.



a) i) The change is speed over 20ºC is 12 m/s. The constant of variation is 12 0 620

.= . Let v

represent the speed, in metres per second, and T represent the temperature, in degrees Celsius.

( )0 6 3310 6 3313

049

3v . T

.= +

= +

=

The speed of sound at 30ºC is 349 m/s. ii) ( )300 6 331

313v .= +

=

−

The speed of sound at –30ºC is 313 m/s. b) ( )100 6 331

325v .= +

=

−

The speed of sound at –10ºC is 325 m/s.

The time required for the sound to travel from Jenny to the wall of the canyon is 1 42. , or 0.7 s.

325 0 7227 5

.d vt

.

== ×=

The wall of the canyon is 227.5 m from Jenny.


Chapter 5 Section 2 Question 14 Page 253 a) b) In each case, C is the charge as a percent and t is the time, in hours.

From 0 to 20 h, the constant of variation is 94 92 0 45 0

.−=

−. The equation is . 0 4 92C . t= +

From 20 to 35 h, there is no change. The equation is 100C = .

For 35 h and more, the constant of variation is 90 95 140 35

−= −

−. The equation is . 135C t= − +

c) i) ( )120 4 92

96 8C .

.= +

= The charge remaining after 12 h was 96.8%. ii) The charge remaining after 26 h was 100%. iii) ( ) 13

4 71 5

6C = − +

= The charge remaining after 71 h was 64%.


Chapter 5 Section 3 Slope Chapter 5 Section 3 Question 1 Page 259

a) riserun350 6

m

.

=

=

=

The slope is 0.6.

b) riserun4 43 21 375

m

.

..

=

=

=

The slope is 1.375. Chapter 5 Section 3 Question 2 Page 259

riserun2 51520 02

m

.

.

=

=

The slope, to the nearest hundredth, is 0.02. Chapter 5 Section 3 Question 3 Page 259

riserun1 48

0 175

m

.

.

=

=

=

The slope of the ramp is 0.175, which does not satisfy the safety regulation of no more than 0.08.



a) riserun350 6

m

.

=

=

=

The slope is 0.6.

b) riserun

350 6

m

.

=

−=

= −

The slope is –0.6. Chapter 5 Section 3 Question 5 Page 259

a) ABriserun13

m =

=

b) CDriserun361 or 0.52

m =

=

=

c) EFriserun

522 5

m

.

=

−=

= −

d) GHriserun050

m =

=

=

e) IJriserun70

Undefined

m =

=

f) KLriserun

250 4

m

.

=

−=

= −


Chapter 5 Section 3 Question 6 Page 260 Answers for the coordinates of point B will vary. A sample answer is shown. Chapter 5 Section 3 Question 7 Page 260

If the slope is 34

− , the run is 4 and the rise is –3. Possible coordinates for B are

(6 + 4, –2 + (–3)), or (10, –5). Answers will vary.



a) riserun560 83

m

.

=

=

=

The slope is . This is outside the safety range of 0.58 to 0.70. 0 83.

b) riserun24280 86

m

.

=

=

The slope is about 0.86. This is outside the safety range of 0.58 to 0.70.


Chapter 5 Section 3 Question 9 Page 260 Answers will vary. Sample answers are shown.

a) If the slope is 23

, the run is 3 and the rise is 2. Possible coordinates for B are

( ) (2 3 5 2 or 1 7, , )− + + .

b) If the slope is 23

− , the run is 3 and the rise is –2. Possible coordinates for B are

( ) ( )2 3 5 2 or 1 3, ,− + − . c) If the slope is , the run is 1 and the rise is 4. Possible coordinates for B are 4( ) ( )2 1 5 4 or 1 9, ,− + + − .

) .

) .

) .

d) If the slope is , the run is 1 and the rise is –3. Possible coordinates for B are 3−( ) (2 1 5 3 or 1 2, ,− + − − e) If the slope is 0 , the run is 1 and the rise is 0. Possible coordinates for B are ( ) (2 1 5 0 or 1 5, ,− + + − f) If the slope is undefined, the line is vertical. Possible coordinates for B are ( ) (2 0 5 1 or 2 6, ,− + + − Chapter 5 Section 3 Question 10 Page 260 Let b represent the length of the vertical brace. First brace: b:1 = 3:5 b = 0.6 m. Second brace: b:2 = 3:5 b = 1.2 m. Third brace: b:3 = 3:5 b = 1.8 m. Fourth brace: b:4 = 3:5 b = 2.4 m.



a) riserun21

5000424 2

m

.. %

=

=

==

The grade of the road is 4.2%. b) Let y represent the required rise.

0 03600

600 0 03 600600

18

y.

y.

y

=

× = ×

=

The road must rise 18 m over a run of 600 m to have a grade of 3%.



a) i) riserun380 375

m

.

=

=

=

The slope of 0.375 is more than 3 0 2512

.= but less than 6 0 512

.= . The pitch is medium.

ii) riserun9

150 6

m

.

=

=

=

The slope of 0.6 is more than 6 0 512

.= . The pitch is steep.

b) If the roof is 10 m wide, the run is 5 m. Let y represent the height.

512 5560 60

12 525 1225 1212 122 1

y

y

yy

. y

=

× = ×

=

=

The height of the roof is about 2.1 m. Chapter 5 Section 3 Question 13 Page 261 The two ramps have the same slope. If the rise is double, the run must be doubled as well. Otherwise, the slopes will be different.


Chapter 5 Section 3 Question 14 Page 261 The rise is 52 – 41 = 11 m. The run is 7 m.

riserun117

1 6

m

.

=

=

The slope is about 1.6. Chapter 5 Section 3 Question 15 Page 261 The rise is 8 m. Let x represent the run.

86 3

86 3

6 3 86 3 86 3 6 3

1 27

.x

x . xx

. x. x. .

x .

=

× = ×

=

=

The maximum distance from the foot of the ladder to the wall is about 1.27 m.

89 5

89 5

9 5 89 5 89 5 9 5

0 84

.x

x . xx

. x. x. .

x .

=

× = ×

=

=

The minimum distance from the foot of the ladder to the wall is about 0.84 m.


Chapter 5 Section 3 Question 16 Page 261 The run is 115 m for a rise of 147 m.

riserun1471151 3

m

.

=

=

The slope is 1.3, which is almost twice as steep as a staircase with a slope of 0.7. Chapter 5 Section 3 Question 17 Page 262 The run is 27.5 m for a rise of 18 m.

riserun18

27 50 65

m

..

=

=

The slope is 0.65, which is about half as steep as the pyramid of Cheops with a slope of 1.3.


Chapter 5 Section 3 Question 18 Page 262 Let x represent the horizontal run.

100 09

100 09

0 09 100 09 100 09 0 09

111 1

.x

x . xx

. x. x. .

x .

=

× = ×

=

=

For a run of more than 111.1 m, the hill is easy.

100 18

100 18

0 18 100 18 100 18 0 18

55 6

.x

x . xx

. x. x. .

x .

=

× = ×

=

=

For a run of 55.6 m to 111.1 m, the hill is intermediate. For a run of less than 55.6 m, the hill is difficult. Chapter 5 Section 3 Question 19 Page 262 Let x represent the run of the trail.

2 2 2

2

2

2

900 6400

7300

30 80

730085 4

x

x

x

xx .

= +

= +

=

=

The hiking trail has a rise of 12 m and a run of about 85.4 m.

riserun12

85 40 14

m

..

=

The slope of the trail is about 0.14.


Chapter 5 Section 3 Question 20 Page 262 From the diagram, each side of the hexagon measures 4 units. Let y represent the rise. The run is 2 units.

2 2 2

2

2

2

2

16 4

16 4 4 4

12

123 4

4

6

2 y

y

y

y

y. y

= +

= +

− = + −

=

=

−

riserun

3 462

1 73

m

.

.

=

−=

= −

The slope is –1.73. Chapter 5 Section 3 Question 21 Page 263 a) Answers will vary. A sample answer is shown. Suppose that one set of stairs has a slope of 0.62 and another has a slope of 0.70. Both sets of stairs are safe, but the set of stairs with the more gradual slope is safer. b) Answers will vary. Chapter 5 Section 3 Question 22 Page 263 Answers will vary. A sample answer is shown. Suppose that there are 5 switchbacks. There needs to be an odd number of switchbacks for the train to end up going in the correct direction. If the run is 1 km, then the slope of each switchback would be 50 ÷ 1000 = 0.05 or 5%, which is less than 7%, as required.


Chapter 5 Section 3 Question 23 Page 263 The base of the triangle measures 6 units. Use the formula for the area of a triangle to find the height.

1122112 62

12 312 33 34

bh

h

hh

h

=

= × ×

=

=

=

riserun4623

m =

=

=

Answer D.


Chapter 5 Section 4 Slope as a Rate of Change Chapter 5 Section 4 Question 1 Page 268

riserun375

7 4

m

.

=

=

=

The rate of change is 7.4 L/min. Chapter 5 Section 4 Question 2 Page 268

riserun7200

24300

m =

=

=

The rate of change is 300 L/h. Chapter 5 Section 4 Question 3 Page 268

riserun1800

3060

m =

=

=

The rate of change is 60 flaps/s. Chapter 5 Section 4 Question 4 Page 268

a) riserun0 268 02683 25

m

.

=

−=

−−

=

= −

The slope is –3.25. b) The height decreases at a rate of 3.25 m/s.



a) riserun

13 252200 200

1220000 006

m

.

=

−=

−−

=

= −

The slope is –0.006. b) The temperature decreases by 0.006°C/m. Chapter 5 Section 4 Question 6 Page 268 The rise is 1.78 – 1.45 = 0.33, and the run is 2006 – 2003 = 3.

riserun0 33

30 11

m

.

.

=

=

=

The rate of change is 11¢/year. Chapter 5 Section 4 Question 7 Page 268

riserun16 261 014610 23

m

.

=

−=

−

=

The rate of change is 0.23 cm/day.


Chapter 5 Section 4 Question 8 Page 269 a) b) Use the points (0, 52 000) and (63, 214 000).

riserun214 000 52 000

63162 000

632571

m =

−=

=

The rate of change is about 2571 downloads/day. c) The software is popular. The number of downloads continues to increase.


Chapter 5 Section 4 Question 9 Page 269 a) b) Use the points (1. 4) and (6, 24).

riserun24 46 1

205

4

m =

−=

−

=

=

The slope is 4. c) The rate of change is 4 toothpicks/diagram.


Chapter 5 Section 4 Question 10 Page 269 Helen is 4 cm taller than John at age 12. John grows one more cm per year than Helen. Helen and John can expect their heights to be the same in 4 years, at age 16. Chapter 5 Section 4 Question 11 Page 269 a)

10 000

20 000

30 000 b) If two hoses are used, the graph will be steeper. It will have twice the slope.


Chapter 5 Section 4 Question 12 Page 269 a) b) Use the points (25, 39) and (75, 117).

riserun117 3975 2578501 56

m

.

=

−=

−

=

=

The rate of change is 1.56 L/m2. c) The amount of water need for a floor area of 140 m2 is 1.56 × 140, or 218.4 L.

If the fire truck is pumping water at a rate of 200 L/min, the time required is 218 4200

. , or about

1.1 min.



b) If it takes 8 s to fill the balloon to 2.5 L, it will take 10 42 5.

= times as long to fill to 10 L, or

32 s. Chapter 5 Section 4 Question 14 Page 270

a) Car A: riserun360

660

m =

=

=

Car B: riserun4805

96

m =

=

=

Car A has a speed of 60 km/h, while car B has a speed of 96 km/h. Car B is faster by 36 km/h. b) The point of intersection of the two lines represents the time at which the two cars have travelled the same distance. If they are travelling in the same direction, it is the time at which Car B passes Car A.


Chapter 5 Section 4 Question 15 Page 270 a) The graph is shown. b) The rate of change was relatively constant from 1990 to 2000. c) The rate of change was different from 2000 to 2005. The rate of change increased. Chapter 5 Section 4 Question 16 Page 270 a) In one minute, the diver will use 15 × 0.002, or 0.03 m3 of air.

b) At this rate, the air will last 2 60 03

..

, or about 87 min.

c) The diver is using the air twice as fast. It will last 87 43 52

.= min.

d) The diver is using the air five times as fast. It will last 875

, or about 17 min.

Chapter 5 Section 4 Question 17 Page 271 a) The rate of change is not constant over the 10-year period. b) Answers will vary. A sample answer is shown. The rates of change are large because the number of jobs increased by about 4300, or 11%, which is a significant amount.


Chapter 5 Section 4 Question 18 Page 271 Solutions for the Achievement Checks are shown in the Teacher's Resource. Chapter 5 Section 4 Question 19 Page 271

126.05

a) b) c) The graph is decreasing. It is curved because the rate of change changes at each interval.


Chapter 5 Section 4 Question 20 Page 271 From 0 min to 100 min:

riserun35 0

100 035

1000 35

m

.

=

−=

−

=

=

From 0 to 100 min, the charge is 35¢/min. From 100 min to 200 min:

riserun60 35

200 10025

1000 25

m

.

=

−=

−

=

=

From 100 min to 200 min, the charge is 25¢/min. For 200 min to 1000 min:

riserun220 60

1000 2001608000 20

m

.

=

−=

−

=

=

For 200 min to 1000 min the charge is 20¢/min.


Chapter 5 Section 5 First Differences Chapter 5 Section 5 Question 1 Page 275 a) The relation is linear. The highest power of x is 1. b) The relation is linear. The highest power of x is 1. c) The relation is non-linear. The highest power of x is 2. d) The relation is non-linear. x is used as an exponent. e) The relation is linear. The highest power of x is 1. f) The relation is non-linear. x appears in the denominator. Chapter 5 Section 5 Question 2 Page 276 a) b) The first differences are not constant. The first differences are constant. The relation is non-linear. The relation is linear. c) d) The first differences are not constant. The first differences are constant. The relation is non-linear. The relation is linear.


Chapter 5 Section 5 Question 3 Page 276 a) The first differences are constant. The relation is linear. b) The first differences are not constant. The relation is non-linear. Chapter 5 Section 5 Question 4 Page 276 a) The first differences are constant. The relation is linear. Let h represent the number of houses, and let S represent the number of segments.

5 1S h= + The seventh step results in 36 segments. b) The first differences are not constant. The relation is non-linear.


Chapter 5 Section 5 Question 5 Page 277 a) The first differences are constant. The relation is linear. Let c represent the number of circles, and let I represent the number of intersection points.

2 2I c= − The seventh step results in 12 intersection points. b) The first differences are not constant. The relation is non-linear. Chapter 5 Section 5 Question 6 Page 277 a) b) The first differences are constant. The relation is linear. c) Let d represent the diagram number, and let T represent the number of toothpicks.

3 1T d= + d) The tenth step results in 31 toothpicks.


Chapter 5 Section 5 Question 7 Page 277 a) b) The first differences are constant. The relation is linear. c) To obtain the wet area, multiply the height by 16. For a height of 50 cm, the wet area is 50 × 16, or 800 cm2. Chapter 5 Section 5 Question 8 Page 278 a) b) The first differences are not constant. The relation is non-linear.


Chapter 5 Section 5 Question 9 Page 278 The first differences are not constant. The relation is non-linear. The points do not fall on a straight line. The relation is non-linear.


Chapter 5 Section 5 Question 10 Page 278 a) b) L3 contains the first differences. The first differences increase by adding 1. Let C represent the number of circles, and let n represent the figure number. Try ( )1C n n= + . This creates the sequence 2, 6, 12, 20,… . Note that each number is double the desired sequence.

Try (1 12

C n n= + ) . This creates the desired sequence.


Chapter 5 Section 6 Connecting Variation, Slope, and First Differences Chapter 5 Section 6 Question 1 Page 284 a) Use (x1, y1) = (–2, 1) and (x2, y2) = (4, 16).

( )

2 1

2 1

16 14 215652

y ymx x−

=−

=−

=

=

−−

b) From the graph, the vertical intercept is 6.

c) The equation of the relation is 5 62

y x= + .

Chapter 5 Section 6 Question 2 Page 284 a) Use (x1, y1) = (–3, –5) and (x2, y2) = (6, 7).

( )( )

2 1

2 1

12

4

567

3

9

3

y ymx x−

=−

−

=

−

=

=−−

b) From the graph, the vertical intercept is –1.

c) The equation of the relation is 4 13

y x= − .



riserun842

m =

=

=

The slope is 2. b)

riserun

1243

m =

−=

= −

The slope is –3.


c)

riserun

6432

m =

−=

= −

The slope is 32

− .

2.0

d)

riserun2 04

0 5

m

.

.

=

=

=

The slope is 0.5.



) The graph is shown.

) Each time the value of increases by 1, the value f y increases by 3. The raph is a straight line that oes not pass through , 0). This is a partial

ariation.

a bxogd(0v ) c

riserun

m =

12=

43

The slope is 3. The vertical intercept is 2. The equation is

=

3 2y x= + .



oes not pass through , 0). This is a partial iation.

c)

a) The graph is shown. b) Each time the value of x increases by 2, the value of y increases by 5. The graph is a straight line that d(0var

riserun2085

m =

=

2=

The slope is 5 . 2

rcept is 16.

The equation is

The vertical inte

5 162

y x= + .


Chapter 5 Section 6 Question 6 Page 285 Let C represent the cost, and r represent the number of rooms.

200 400C r= + Chapter 5 Section 6 Question 7 Page 285 a) Click here to load the Fathom® file. b) Use (x1, y1) = (0.5, 5.75) and (x2, y2) = (2.5, 8.75).

2 1

2 1

8 75 5 752 5 0 5. .

. .−−

3 002 0

1 5

y ymx x

..

.

−=

−

=

=

=

The slope is 1.5. This represents the variable cost of $1.50 per km. The vertical intercept is 5.00. This represents the fixed cost of $5.00. c) This is a partial variation. The graph is a straight line that does not pass through (0, 0). d) Let C represent the cost, and d represent the number of kilometres. The equation is . 1 5 5 00C . d .= +


Chapter 5 Section 6 Question 8 Page 285 Each second, the scuba diver swims 1 m toward the surface of the water. The rise is 20, and the run is 20.

riserun20201

m =

=

=

The slope is 1, and the vertical intercept is –50. Let D represent the depth, in metres, and t represent the time, in seconds.

50D t= −


Chapter 5 Section 6 Question 9 Page 286 a) Since the variation is direct, the graph passes through (0, 0). Use (x1, y1) = (0, 0) and (x2, y2) = (4, 9).

2 1

2 1

9 004

94

y ymx x−

=−−

=−

=

The slope is 94

. The vertical intercept is 0.

b) The equation is 94

y x= .

c)


Chapter 5 Section 6 Question 10 Page 286 a) Use (x1, y1) = (0, 5) and (x2, y2) = (6, 8).

2 1

2 1

8 56 03612

y ymx x−

=−

=

=

=

−−

The slope is 12

. The vertical intercept is 5.

b) The equation is 1 52

y x= + .

c)


Chapter 5 Section 6 Question 11 Page 286 y varies partially with x. As the value of x increases by 3, the value of y decreases by 7.

riserun

21973

m =

−=

= −

The slope is 73

− . The vertical intercept is –5.

The equation is 7 53

y x= − − .


4 3y x= − y varies partially with x. As the value of x increases by 1, the value of y increases by 4.


Chapter 5 Section 6 Question 13 Page 286 a) The rate of change is the same for each succeeding pair of data points. The relation is linear. b)

c) The rate of change from part a) is the slope of the graph. The slope is –0.25 or 14

− . The slope

is constant. It represents the rate of change of the volume of water in the pool. Water is draining out at a rate of 0.25 kL/min. d) Let V represent the volume of water in the pool, in kilolitres, and t represent the time, in minutes. The vertical intercept is 50.

0 25 50V . t= − + e) ( )0 25 50

1 560

5 035

V .= − +

= − +=

The volume of water after 60 min is 35 kL.


Chapter 5 Section 6 Question 14 Page 287 Solutions for Achievement Checks are shown in the Teacher's Resource. Chapter 5 Section 6 Question 15 Page 287 a) Graph the mass versus dosage data. Extend the graph to determine the vertical intercept. The vertical intercept is 10. Use (x1, y1) = (40, 30) and (x2, y2) = (120, 70).

2 1

2 1

70 30120 4040

180

2

y ymx x−

=−−

=

=

=

−

Let D represent the dosage, in milligrams, and let m represent the mass of the patient, in

kiolgrams. The equation is 1 102

D m= +

b) 11 10 10211 10 1 10 102

11 1120

D . m

. m .

m

⎛ ⎞= +⎜ ⎟⎝ ⎠

= × + ×

= +

c) The graphs are shown. The graph of the maximum dosage has a vertical intercept of 11, which is 1 higher than the vertical intercept of the recommended dosage, 10. The maximum dosage graph rises more steeply.


Chapter 5 Section 6 Question 16 Page 287 Assume that the percent commission is constant. Use (x1, y1) = (15 000, 1300) and (x2, y2) = (34 000, 1680).

2 1

2 1

1680 130034 000 15 000

1300 0

38019 0000 02

y ymx x

.

−=

−−

=−

=

=

15

001000

00

Base Salary

.= × += +

− = + −=

The equation is S . Use the first pair of numbers in the table to find the Base Salary.

alary 0 02 Sales Base Salary.= × +

0 02 Base Salary

1300 300 Base Salary1300 300 300 Base Salary 3

The base salary is $1000 per month, and the rate of commission on sales is 0.02 or 2%.


Chapter 5 Review Chapter 5 Review Question 1 Page 288 a) b) c) Let P represent the pay, in dollars, and let t represent the time worked, in hours.

9P t= Chapter 5 Review Question 2 Page 288

a) The constant of variation is 144 961 5.

= . This represent a speed of 96 km/h.

96d t=

b) 96

300 9696 9

3 5

300

612

tt

. t

=

=

=

The time required to reach their destination is 3 h 7 min 30 s.


Chapter 5 Review Question 3 Page 288 a) This is a direct variation. The volume of soup varies directly with the volume of water used to prepare it. b) c) If John uses 2.8 L of water to make 3.0 L of soup, the graph becomes less steep. Chapter 5 Review Question 4 Page 288 a) b) The initial value of y is 4. When x changes by 1, y changes by 3. The constant of variation is 3. c) 3 4y x= + d) This graph is a straight lien that starts at (0, 4) and rises upward to the right with a slope of 3.


Chapter 5 Review Question 5 Page 288 a) The variation is neither direct nor partial. It is not a straight line. b) This is a partial variation. It is a straight line that does not pass through (0, 0). c) This is a direct variation. It is a straight line that passes through (0, 0). d) This is a partial variation. It is a straight line that does not pass through (0, 0). Chapter 5 Review Question 6 Page 288 a) The fixed cost is $500. The variable cost is $0.15 times the number of flyers printed. b) Let C represent the cost, and let f represent the number of flyers printed.

0 15 500C . f= + c) ( )0 15 500

755

50057

0

5

0C .= +

= +=

The cost of printing 500 flyers is $575. Chapter 5 Review Question 7 Page 288

a) riserun0 262 0

0 13

m

..

.

=

=

=

The slope is 0.13.

b) riserun45321 4

m

.

=

=

The slope is about 1.4.


Chapter 5 Review Question 8 Page 288

a) ABriserun14

m =

=

The slope of segment AB is 14

.

b) CDriserun54

m =

=

The slope of segment CD is 54

.

c) EFriserun

34

m =

−=

The slope of segment EF is 34

− .


Chapter 5 Review Question 9 Page 289 Chapter 5 Review Question 10 Page 289

riserun2

0 45

m

.

=

=

=

The slope of the ladder is 5. It is not within the safe range of 6.3 to 9.5.


Chapter 5 Review Question 11 Page 289

riserun24300 8

m

.

=

=

=

Walking burns 0.8 kJ/min.

riserun36301 2

m

.

=

=

=

Cycling burns 1.2 kJ/min.

riserun48301 6

m

.

=

=

=

Swimming burns 1.6 kJ/min.

riserun84302 8

m

.

=

=

=

Playing basketball burns 2.8 kJ/min.


Chapter 5 Review Question 12 Page 289 Use the points (12, 45) and (17, 106).

riserun106 4517 12615

12 2

m

.

=

−=

−

=

=

The slope of the graph is 12.2. Hair grows at a rate of 12.2 cm/year. Chapter 5 Review Question 13 Page 289 a) b) The first differences are constant. The first differences are not constant. The relation is linear. The relation is non-linear. Chapter 5 Review Question 14 Page 289

The first differences are constant. The relation is linear.


Chapter 5 Review Question 15 Page 289 a) The first differences are constant. The relation is linear. b) Use (x1, y1) = (0, 2) and (x2, y2) = (4, 14).

2 1

2 1

14 24 0

124

3

y ymx x−

=−−

=−

=

=

The slope is 3. c) The vertical intercept is 2. The equation is 3 2y x= + . d)


Chapter 5 Review Question 16 Page 289 a) The first differences are constant. The relation is linear. b) c) Use (x1, y1) = (0, 9.0) and (x2, y2) = (5, 7.0).

2 1

2 1

7 0 9 05 0

.

2 050 4

y ymx x

.

.

−=

−

=−

−=

= −

.−

The slope is –0.4. This means the propane is used up at a rate of 0.4 kg/h. The vertical intercept is 9.0. This is the initial amount of propane, in kilograms. d) Let m represent the mass, in kilograms, and let t represent the time, in hours. The equation is

. 0 4 9 0m . t .= − +


Chapter 5 Chapter Test Chapter 5 Chapter Test Question 1 Page 290 B and D are non-linear. A is a direct variation. The only partial variation is answer C. Chapter 5 Chapter Test Question 2 Page 290

The constant of variation is 150 1001 5.

= . Answer A.

Chapter 5 Chapter Test Question 3 Page 290

riserun340 75

m

.

=

=

=

The slope is 0.75. Answer C. Chapter 5 Chapter Test Question 4 Page 290 Non-linear relations do not have constant first differences. Linear relations have constant first differences. Answer C is false. Chapter 5 Chapter Test Question 5 Page 290

The constant of variation is 43 50 0 8750. .= . Since the variation is direct, the correct answer is D.


Chapter 5 Chapter Test Question 6 Page 290 a) Use (x1, y1) = (–4, 7) and (x2, y2) = (2, –2).

( )

2 1

2 1

2 72 4

96

23

y ymx x−

=−

=−

−=

= −

−−−

The slope is 32

− .

b) The vertical intercept is 1.

c) The equation for the relation is 3 12

y x= − + .

Chapter 5 Chapter Test Question 7 Page 290

a) 685 342 52 0

..

=

The rate of change is 342.5 m/s. So, the slope is 342.5. Let d represent the distance, in metres, and let t represent the time, in seconds. The equation for the relation is . 342 5d .= t b)


Chapter 5 Chapter Test Question 8 Page 290 The first differences are constant. The relation is linear. Chapter 5 Chapter Test Question 9 Page 290 a) Let P represent the price charged, in dollars, and let t represent the time, in hours. The equation is . 50 60P t= + b) ( )50 60

13

75

5 60235

P .= +

= +=

The total cost of a repair that takes 3.5 h is $235. c) If the hourly cost changed to $45, the equation would become 45 60P t= + .


Chapter 5 Chapter Test Question 10 Page 291 a) Use (x1, y1) = (0, 8000) and (x2, y2) = (150, 17 000).

2 1

2 1

17 000 8000150 0

−

9000150

60

y ymx x−

=−

=−

=

=

.

The rate of change is $60/page. This is the slope of the graph. b) The equation is . 60 8000C p= + c) If the base cost changed to $9000, the vertical intercept would be 9000, and the equation would be . 60 9000C p= + d) The new cost per page would be 1 08 60 64 8. × = . The new equation would be

. 64 8 8000C . p= +


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Get Ready for Grade 8panchbhaya.weebly.com/uploads/1/3/7/0/13701351/pom9-solutions-chapter... · Chapter 5 Get Ready Chapter 5 Get Ready Question 1 Page 236 a) 3 4 − is not equivalent